A certain test preparation course is designed to help students improve their scores on the USMLE exam. A mock exam is given at the beginnin

Question

A certain test preparation course is designed to help students improve their scores on the USMLE exam. A mock exam is given at the beginning and end of the course to determine the effectiveness of the course. The following measurements are the net change in 3 students’ scores on the exam after completing the course: 15,20,18 Using these data, construct a 90% confidence interval for the average net change in a student’s score after completing the course. Assume the population is approximately normal. Step 3 of 4 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

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Maria 3 weeks 2022-01-01T18:35:57+00:00 1 Answer 0 views 0

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  1. Ava
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    2022-01-01T18:37:09+00:00

    Answer:

    17.67-2.92\frac{2.517}{\sqrt{3}}=13.427    

    17.67+2.92\frac{2.517}{\sqrt{3}}=21.913    

    So on this case the 90% confidence interval would be given by (13.427;21.913)    

    Step-by-step explanation:

    Previous concepts

    A confidence interval is “a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval”.

    The margin of error is the range of values below and above the sample statistic in a confidence interval.

    Normal distribution, is a “probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean”.

    \bar X represent the sample mean for the sample  

    \mu population mean (variable of interest)

    s represent the sample standard deviation

    n represent the sample size  

    Solution to the problem

    The confidence interval for the mean is given by the following formula:

    \bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

    In order to calculate the mean and the sample deviation we can use the following formulas:  

    \bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

    s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

    The mean calculated for this case is \bar X=17.67

    The sample deviation calculated s=2.517

    In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

    df=n-1=3-1=2

    Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: “=-T.INV(0.05,2)”.And we see that t_{\alpha/2}=2.92

    Now we have everything in order to replace into formula (1):

    17.67-2.92\frac{2.517}{\sqrt{3}}=13.427    

    17.67+2.92\frac{2.517}{\sqrt{3}}=21.913    

    So on this case the 90% confidence interval would be given by (13.427;21.913)    

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